Problem: Simplify the following expression and state the condition under which the simplification is valid. $t = \dfrac{a^3 - 25a}{-5a^2 - 20a + 25}$
Answer: First factor out the greatest common factors in the numerator and in the denominator. $ t = \dfrac {a(a^2 - 25)} {-5(a^2 + 4a - 5)} $ $ t = -\dfrac{a}{5} \cdot \dfrac{a^2 - 25}{a^2 + 4a - 5} $ Next factor the numerator and denominator. $ t = - \dfrac{a}{5} \cdot \dfrac{(a + 5)(a - 5)}{(a + 5)(a - 1)}$ Assuming $a \neq -5$ , we can cancel the $a + 5$ $ t = - \dfrac{a}{5} \cdot \dfrac{a - 5}{a - 1}$ Therefore: $ t = \dfrac{ -a(a - 5)}{ 5(a - 1)}$, $a \neq -5$